Find the function whose tangent line has a slope x ln x12 fo

Find the function whose tangent line has a slope x ln (x)^1/2 for each value x>0 and whose graph
passes through point (2,-3)

f(2)=-3 f\'(x)=x*ln(sqrt(x))=(x/2)*ln(x) by logarithmic rules. u=ln(x), du= dx/x dv=(x/2), v=x^2/4 f=int(f\' dx)=uv-int(vdu) f(x)=(x^2/4)*ln(x)-int(x^2/4*(dx/x)) f(x)=x^2*ln(x)/4-(1/4)*int(x dx) f(x)=x^2*ln(x)/4-(1/8)*x^2+C f(2)=-3=(4)*ln(4)/4-(1/8)*4+C -3=ln(4)-(1/2)+C C= -(5/2+ln(4)) f(x)=x^2*ln(x)/4-(x^2/8)-(5/2+ln(4)) Assuming I was able to read your f\'(x) correctly, since it is slightly ambiguous.

$Find the function whose tangent line has a slope x ln (x)^1/2 for each value x>0 and whose graph passes through point (2,-3)Solution f(2)=-3 f\'(x)=x*ln(sqrt$

Find the function whose tangent line has a slope x ln x12 fo

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